Áp dụng BĐT $$\frac{1}{x}+\frac{1}{y}\ge \frac{4}{x+y}$$

\(Q=1964\left(\frac{1}{p-a}+\frac{1}{p-b}\right)+15\left(\frac{1}{p-b}+\frac{1}{p-c}\right)+10\left(\frac{1}{p-a}+\frac{1}{p-c}\right)\)

\(\ge1964\cdot\frac{4}{2p-\left(a+b\right)}+15\cdot\frac{4}{2p-\left(b+c\right)}+10\cdot\frac{4}{2p-\left(c+a\right)}\)

\(=4\left(\frac{15}{a}+\frac{10}{b}+\frac{1964}{c}\right)=4\cdot2006=8024\)

Xảy ra khi \(a=b=c=\frac{117}{118}\)

tìm GTLN trừ GTNN hay GTLN riêng và GTNN riêng

riêng nha bạn

\(\frac{x^3-y^3}{\left(x-y\right)^3}=\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)^3}=\frac{x^2+xy+y^2}{x^2-2xy+y^2}\)

\(=\frac{\frac{1}{4}\left(x^2-2xy+y^2\right)+\frac{3}{4}\left(x^2+2xy+y^2\right)}{x^2-2xy+y^2}\)

\(=\frac{1}{4}+\frac{3}{4}.\frac{\left(x+y\right)^2}{\left(x-y\right)^2}\ge\frac{1}{4}\)

\(P=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

\(=\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\left(1\right)\)

Áp dụng BĐT AM-GM ta có: :

\(\frac{a}{a^2+b^2+c^2}+9a\left(a^2+b^2+c^2\right)\ge2\sqrt{9a^2}=6a\)

Tương tự cho 2 BĐT còn lại ta cũng có: 

\(\frac{b}{a^2+b^2+c^2}+9b\left(a^2+b^2+c^2\right)\ge6b;\frac{c}{a^2+b^2+c^2}+9c\left(a^2+b^2+c^2\right)\ge6c\)

\(\Rightarrow\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}+9\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge6\left(a+b+c\right)\)

Theo BĐT Cauchy-Schwarz thì:

\(9\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge9\cdot\frac{\left(a+b+c\right)^2}{3}\cdot\left(a+b+c\right)=3\)

\(\Rightarrow\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}\ge6-3=3\)

Và \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{9}{ab+bc+ca}\ge\frac{9}{\frac{\left(a+b+c\right)^2}{3}}=27\)

Khi đó nhìn vào \(\left(1\right)\) thấy \(P\ge27+3=30\)

Xảy ra khi \(a=b=c=\frac{1}{3}\)

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